# Mathematics Questions and Answers – Methods of Integration-1

«
»

This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Methods of Integration-1”.

1. Find ∫7 cos⁡mx dx.
a) $$\frac{7 \,sin⁡mx}{x}+C$$
b) $$\frac{7 \,sin⁡mx}{m}+C$$
c) $$\frac{sin⁡mx}{x}+C$$
d) $$\frac{sin⁡x}{m}+C$$

Explanation: Using Integration by Substitution, Let xm=t
Differentiating w.r.t x, we get
mdx=dt
∴$$\int 7 \,cos⁡mx \,dx=\int \frac{(7 cos⁡t)}{m} dt$$
=$$\frac{7}{m} \int cos⁡t \,dt=\frac{7}{m} (sin⁡t)+C$$
Replacing t with mx again we get,
$$\int 7 \,cos⁡mx \,dx=\frac{7 \,sin⁡mx}{m}+C$$

2. Integrate $$3x^2 (cos⁡x^3+8)$$.
a) $$sin⁡x^3-8x^3+C$$
b) $$sin⁡x^3+8x^3+C$$
c) –$$sin⁡x^3+8x^3+C$$
d) $$sin⁡x^3-x^3+C$$

Explanation: By using the method of integration by substitution,
Let x3=t
Differentiating w.r.t x, we get
3x2 dx=dt
$$\int 3x^2 \,(cos⁡x^3+8) \,dx=\int (cos⁡t+8)dt$$
$$\int (cos⁡t+8) dt=sin⁡t+8t$$
Replacing t with x3,we get
$$\int 3x^2 (cos⁡x^3+8) dx=sin⁡x^3+8x^3+C$$

3. Find $$\int 6x(x^2+6)dx$$.
a) $$\frac{3x^4}{2}+18x^2+C$$
b) $$\frac{3x^4}{2}-18x+C$$
c) $$\frac{3x^4}{2}-18x^2+C$$
d) $$\frac{3x^4}{2}+x^2+C$$

Explanation: Let x2=t
Differentiating w.r.t x, we get
2x dx=dt
$$\int 6x(x^2+6)dx=3\int (t+6) dt$$
3$$\int (t+6)dt=3\left (\frac{t^2}{2}+6t\right )=\frac{3t^2}{2}+18t$$
Replacing t with x2
$$\int 6x(x^2+6)dx=\frac{3x^4}{2}+18x^2+C$$

4. Find the integral of $$3e^x+\frac{2(log⁡ x)}{3x}$$.
a) $$3e^x+\frac{1}{3} (x)^2+C$$
b) $$e^x-\frac{8}{3} (log⁡x)^2+C$$
c) $$3e^x-\frac{1}{3} (log⁡x)^2+C$$
d) $$3e^x+\frac{1}{3} (log⁡x)^2+C$$

Explanation: $$\int 3e^x+\frac{2(log⁡x^2)}{3x} dx=3\int e^x dx+\frac{2}{3} \int \frac{log⁡x}{x}$$
Let log⁡x=t
Differentiating w.r.t x, we get
$$\frac{1}{x} dx=dt$$
∴$$\int \frac{log⁡x}{x}=\int \,t \,dt=\frac{t^2}{2}$$
$$\int e^x dx=e^x$$
Replacing t with log⁡x, we get
$$\int 3e^x+\frac{2(log⁡x^2)}{3x} dx=3e^x+\frac{1}{3} (log⁡x)^2+C$$

5. Find $$\int \frac{e^{-cot^{-1}⁡x}}{1+x^2}$$.
a) $$e^{-cot^{-1}⁡x}+C$$
b) $$e^{-2cot^{-1}⁡x}+C$$
c) $$e^{-tan^{-1}⁡x}+C$$
d) $$e^{-cot^1⁡2x}+C$$

Explanation: Let $$-cot^{-1}⁡x$$=t
Differentiating w.r.t x, we get
–$$\left (-\frac{1}{1+x^2}\right )dx=dt$$
$$\frac{1}{1+x^2} dx=dt$$
$$\int \frac{e^{-cot^{-1}}x}{1+x^2} dx=\int e^t \,dt$$
=et
Replacing t with -cot-1x, we get
$$\int \frac{e^{-cot^{-1}}x}{1+x^2} dx=e^{-cot^{-1}}x+C$$
Participate in Mathematics - Class 12 Certification Contest of the Month Now!

6. Find the integral of $$\frac{5x^4}{\sqrt{x^5+9}}$$.
a) $$\sqrt{x^5+9}$$
b) $$2\sqrt{x^5-9}$$
c) 2(x5+9)
d) $$2\sqrt{x^5+9}$$

Explanation: Let x5+9=t
Differentiating w.r.t x, we get
5x4 dx=dt
$$\int \frac{5x^4}{\sqrt{x^5+9}} dx=\int \frac{dt}{\sqrt{t}}$$
=$$\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=2\sqrt{t}$$
Replacing t with x5+9, we get
$$\int \frac{5x^4}{\sqrt{x^5+9}} dx=2\sqrt{x^5+9}$$.

7. Find $$\int \frac{6 sin⁡\sqrt{x}}{\sqrt{x}} dx$$
a) $$2 \,cos⁡\sqrt{x}+C$$
b) –$$12 \,cos⁡\sqrt{x}+C$$
c) -12 cos⁡x+C
d) 12 cos⁡x+C

Explanation: Let $$\sqrt{x}=t$$
Differentiating w.r.t x,we get
$$\frac{1}{2\sqrt{x}} dx=dt$$
$$\frac{1}{\sqrt{x}} dx=2dt$$
∴$$\int \frac{6 sin⁡\sqrt{x}}{\sqrt{x}} dx=\int \,12 \,sin⁡t \,dt$$
=12(-cos⁡t)=-12 cos⁡t
Replacing t with $$\sqrt{x}$$, we get
$$\int \frac{6 sin⁡\sqrt{x}}{\sqrt{x}} dx=-12 \,cos⁡\sqrt{x}+C$$

8. Find $$\int \frac{20x^3}{1+x^4} dx$$.
a) 5 log⁡(x4)+C
b) -5 log⁡(1+x4)+C
c) 5 log⁡(1+x4)+C
d) log⁡(1+x4)+C

Explanation: Let 1+x4=t
4x3 dx=dt
∴$$\int \frac{20x^3}{1+x^4} dx=5\int \frac{dt}{t}$$
=5 log⁡t
Replacing t with 1+x4, we get
$$\int \frac{20x^3}{1+x^4} dx=5 \,log⁡(1+x^4)+C$$

9. Integrate $$\frac{x^2}{e^{x^3}}$$.
a) –$$\frac{1}{(3e^{x^3})}+C$$
b) $$\frac{1}{3e^{x^3}}+C$$
c) –$$\frac{1}{e^{x^3}}+C$$
d) ex3+C

Explanation: Let x3=t
3x2 dx=dt
x2 dx=dt/3
∴$$\int \frac{x^2}{e^{x^3}} dx=\frac{1}{3} \int \frac{dt}{e^t}$$
=$$\frac{1}{3} \left (-e^{-t}\right )$$
Replacing t with x3, we get
$$\int \frac{x^2}{e^{x^3}} dx=-\frac{1}{3e^{x^3}}+C$$

10. Find $$\int \frac{cos^{-1}x}{\sqrt{1-x^2}} dx$$.
a) $$\frac{(sin^{-1}⁡x)^2}{2}+C$$
b) $$\frac{(cos^{-1}⁡x)^2}{7}+C$$
c) $$\frac{(cos^{-1}⁡x)^2}{2}+C$$
d) –$$\frac{(cos^{-1}⁡x)^2}{2}+C$$

Explanation: Let cos-1⁡x=t
Differentiating w.r.t x, we get
$$\frac{1}{\sqrt{1-x^2}} dx=dt$$
∴$$\int \frac{cos^{-1}⁡x}{\sqrt{1-x^2}} dx=\int t dt$$
=$$\frac{t^2}{2}$$
Replacing t with cos-1x,we get
$$\int \frac{cos^{-1}⁡x}{\sqrt{1-x^2}} dx=\frac{(cos^{-1}⁡x)^2}{2}+C$$

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.